Math 180B Problem Set 3

Transcription

1 Math 180B Problem Set 3 Problem 1. (Exercise 3.1.2) Solution. By the definition of conditional probabilities we have Pr{X 2 = 1, X 3 = 1 X 1 = 0} = Pr{X 3 = 1 X 2 = 1, X 1 = 0} Pr{X 2 = 1 X 1 = 0} = P 1,1 P 0,1 =.6.2 =.12. Note the second equality uses the Markov property. Similarly we have Pr{X 1 = 1, X 2 = 1 X 0 = 0} = Pr{X 2 = 1 X 1 = 1, X 0 = 0} Pr{X 1 = 1 X 0 = 0} = P 1,1 P 0,1 =.6.2 =.12. Problem 2. (Exercise 3.1.5) Solution. Using conditional probabilities repeatedly (as in Eq. 3.8, pp. 80) we find Pr{X 0 = 1, X 1 = 1, X 2 = 0} = p 1 P 1,1 P 1,0 = =.025. To determine the next probability we employ the law of total probability by summing over the events X 0 = 0 and X 0 = 1: Pr{X 1 = 1, X 2 = 1, X 3 = 0} = p 0 P 0,1 P 1,1 P 1,0 + p 1 P 1,1 P 1,1 P 1,0 = = = = = Problem 3. (Problem 3.1.2) Solution. (a) Using conditional probability as above we have Pr{X 0 = 0, X 1 = 0, X 2 = 0} = p 0 P 0,0 P 0,0 = 1 (1 α) (1 α) = (1 α) 2. (b) Since we know that X 0 = 0 with probability one, we must compute the likelihood that X 2 = 0 given that X 0 = 0. For this to happen, either X 1 = 0 or X 1 = 1. Therefore Pr{X 2 = 0 X 0 = 0} = Pr{X 2 = 0, X 1 = 0 X 0 = 0} + Pr{X 2 = 0, X 1 = 1 X 0 = 0} = = P 0,0 P 0,0 + P 0,1 P 1,0 = (1 α) 2 + α 2 = 2α 2 2α

2 Problem 4. (Exercise 3.2.2) Solution. For n = 0 we have that Pr{X 0 = 0 X 0 = 0} = 1. For n 1 the probability Pr{X n = 0 X 0 = 0} is equal to P (n) 0,0 which is equal to the (0, 0) entry of P n (the nth power of the matrix P). Reading off the matrix P we find P 0,0 = 0. For n > 1 we calculate the first few powers of P: 1/2 1/4 1/4 1/4 3/8 3/8 3/8 5/16 5/16 P 2 = 1/4 1/2 1/4, P 3 = 3/8 1/4 3/8, P 4 = 5/16 3/8 5/16 1/4 1/4 1/2 3/8 3/8 1/4 5/16 5/16 3/8 Reading off the (0, 0) entries we find P (2) 0,0 = 1/2, P (3) 0,0 = 1/4, P (4) 0,0 = 3/8. Problem 5. (Exercise 3.2.6) Solution. To compute Pr{X 2 = 0} and Pr{X 3 = 0} we first compute the 2nd and 3rd powers of the transition matrix P: P 2 = , P 3 = By Eq 3.13 (p. 84) we compute Pr{X 2 = 0} by combining the matrix elements in the 0th column of P 2 with the initial distribution X 0 : Pr{X 2 = 0} = p 0 P 2 0,0 + p 1 P 2 1,0 + p 2 P 2 2,0 = =.42. Same idea for X 3 : Pr{X 3 = 0} = p 0 P 3 0,0 + p 1 P 3 1,0 + p 2 P 3 3,0 = =.416. Problem 6. (Problem 3.2.2) Solution. Note that X 5 = 0 given that X 0 = 0 if and only if there are an even number of errors in the first 5 stages of transmission. Since the probability of error at each stage is always α (and at each stage the chance of error is independent of previous stages), the total number of errors follows a binomial distribution with parameters 5 and α. Let Y denote the distribution Binomial(5, α). Then Pr{X 5 = 0 X 0 = 0} = Pr{Y is even } = Pr{Y = 0} + Pr{Y = 2} + Pr{Y = 4} = (1 α) 5 + ( ) 5 α 2 (1 α) ( ) 5 α 4 (1 α). 4 2

3 Problem 7. (Problem 3.2.5) Solution. As expressed in the hint, the event {T > 3} is equivalent to {X 3 = 0 or X 3 = 1}. Therefore, by the definition of conditional probability, Pr{X 3 = 0 X 0 = 0, T > 3} = We have Pr{X 3 = 0 X 0 = 0} Pr{X 3 = 0 X 0 = 0} + Pr{X 3 = 1 X 0 = 0} = P 3 0,0 P 3 0,0 +. P3 0, P 3 = so the probability above comes out to be =.665. Problem 8. (Exercise 3.3.1) Solution. The possible values for X n are 1, 0, 1, 2, 3. One fills out the matrix of transition probabilities entry by entry. For example, if X n = 1 then in the next stage the inventory is restocked and so Pr{X n+1 = 3 X n = 1} = P 1,3 =.4, P 1,2 =.3, P 1,1 =.3, P 1,0 = P 1, 1 = 0. This fills out the 1-row of the transition matrix. The rest of the entries are determined in a similar way. The final transition matrix is Problem 9. (Problem 3.3.2) Solution. Each random variable X n is counting the number of heads or tails out of 3 coins, so the possible values for each variable are 0, 1, 2, 3. Suppose that X n counts heads and X n+1 counts tails. Then X n is equal to the number of heads heads at time n. Thus there are 3 X n tails. To determine the value of X n+1 we toss the X n coins which were heads, and then we look at the total number of tails. For example, if X n = 1 then we only toss one coin. With probability 1/2 we see tails, which means X n+1 = = 3. With probability 1/2 we see heads in which case X n+1 = 2. Thus P 1,2 = P 1,3 =.5 and P 1,0 = P 1,1 = 0. This accounts for the 1-row of the transition matrix. The other entries are 3

4 determined in a similar way. The final matrix is /2 1/ /4 1/2 1/4. 3 1/8 3/8 3/8 1/8 Problem 10. (Problem 3.3.7) Solution. The variables X n are discrete and take the values 0, 1, 2,.... As mentioned in the problem, if X n = 0 then X n+1 is the total operating life of the next component. By assumption the total operating life takes the value k with probability α k. Therefore P 0,k = α k and P 0,0 = 0. Next, note that if X n 0 then X n+1 = X n 1. Therefore for i > 0 and j 0 we have { 0 j i 1 P i,j = 1 j = i 1. These two rules describe all the transition probabilities. Problem 11. Suppose X, Y have a bivariate normal distribution with X N (0, 1) and Y X = x N (ax + b, σ 2 ). a. What is the marginal distribution of Y? b. What is the conditional distribution X Y = y? Solution. First define a new random variable Ŷ := Y b. Answering the above questions for Ŷ will also yield the answers for Y. Note that Ŷ X = x N (ax, σ 2 ), so the mean of Ŷ is proportional to X. To find the marginal and conditional distributions above we must find the correlation of Ŷ with X (denoted ρ) and the variance σ y of Ŷ. Following the notes on bivariate normal distributions, these values are determined by solving ρ σ y σ x = ρσ y = a (1 ρ 2 )σ 2 y = σ 2. To get the first equation we used the fact that σ x = 1. Solving the above equations for ρ and σ y yield ρ = a σ2 + a 2 σ y = σ 2 + a 2. 4

5 Now the joint density function for X and Ŷ depends only on ρ, σ x and σ y (see Eq. 3 in the notes) and the dependence on σ x and σ y is symmetric. Therefore Ŷ N (0, σ 2 y) = N (0, σ 2 + a 2 ) X Ŷ = y N (ρσ x y, (1 ρ 2 )σ 2 a σ x) = N ( y σ 2 + a 2 y, σ 2 σ 2 + a 2 ). Finally we can answer the questions about Y. The marginal distribution for Y is Y = Ŷ + b N (b, σ2 + a 2 ) while the conditional distribution of X given Y = y is (X Y = y) = (X Ŷ = y b) N ( a σ 2 + a 2 (y b), σ 2 σ 2 + a 2 ). 5